$\sqrt[13]{K^{7}}=\frac{23}{37}$

∴ $K^{\frac{7}{13}}=\frac{23}{37}$ ∴ k = $\left ( \frac{23}{37} \right )^{\frac{13}{7}}$

Since \frac{23}{37} < 1, higher powers i.e. power> 1 will have lesser values.

∴ $\frac{23}{37}> \left ( \frac{23}{37} \right )^{\frac{13}{7}}$

Area of shaded region =

(πy² - π3x²) = π(y² - 3x²)

Area of white region = π3x²

∴ π(y² - 3x²) = π3x²

∴ y² = 6x² ∴ ^y⁄_x = √6 > 2

$\theta =\frac{50\times 51}{2}=1275$

φ = $\left ( \frac{100\times 101}{2} \right )-\frac{50\times 51}{2}=3775$

∴ φ - θ = 2500

Col. A

XYZ is an isosceles right angle triangle.

∴ xy = yz = √2

∴ Area = ¹⁄₂ × √2 × √2 = 1

Col. B

PQR is 30° - 60° - 90° Triangle

∴ PQ = 1 QR = √3

Area = ¹⁄₂ × 1 × √3 = .866

¹⁄_x - ¹⁄_y = 1 ∴ $\frac{y-x}{xy}=1$

∴ y - x = xy

If x and y both are positive,

⇒ xy is positive ⇒ y - x = positive

⇒ y > x

However, if either of them is negative,

⇒ xy is negative ⇒ y < x is negative. ⇒ y < x

2⁷ - 2⁵ - 2⁵ = 2⁵(2² - 2¹ - 1)

= 2⁵ = 32

At 4:20 p.m., hour hand will move down towards 5 p.m. Therefore there will be some angle between hour hand and minute hand.

Formula:

(Theoretical distance in two hands × 60 ) + (Time × ¹⁄₂°)

= (0 × 6°) + (20 × ¹⁄₂°) = 10°

Theoretical distance is calculated assuming that hour hand does not move between 4 to 5.

All the corners will have 3 faces painted blue.

∴ Col. A = 8

Col. B The

core of the big cube is formed of 8 small 11 inch cubes which have no face painted

∴ Col. B = 8

Col. A-

$\frac{5 + 10 +15 +20 + 25+ 30+ 35+40+45}{9}=25$

Col. B-

2 × 3 × 5 = 30

P has a positive value and q is zero, (p, q) lies on x axis.

y has a negative value and x is zero, (x, y) lies on y axis.

∴ p + x = p = positive number

q + y = y = negative number